The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$ and having a normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{-1} = \frac{z + 7}{-1}$ is . . . .

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then:

Find the coordinates of the point of intersection of the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ and the plane $x + y - z = 3$.

Let $P_1: 2x + y - z = 3$ and $P_2: x + 2y + z = 2$ be two planes. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ The line of intersection of $P_1$ and $P_2$ has direction ratios $1, -1, 1$.
$(B)$ The line $\frac{3x - 4}{9} = \frac{1 - 3y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_1$ and $P_2$.
$(C)$ The acute angle between $P_1$ and $P_2$ is $60^{\circ}$.
$(D)$ If $P_3$ is the plane passing through the point $(4, 2, -2)$ and perpendicular to the line of intersection of $P_1$ and $P_2$,then the distance of the point $(2, 1, 1)$ from the plane $P_3$ is $\frac{2}{\sqrt{3}}$.

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$ and is perpendicular to the plane $2x+3y+z=5$.